1. The energy and the linear momentum of a distribution of electromagnetic fields in vacuum is given (in SI units) by. d 3 r( E 2 +c 2 B 2 ), (1) 2

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1 Physics 4 Solution Set Winter 7. The energy and the linear momentum of a distribution of electromagnetic fields in vacuum is given (in SI units) by U = ǫ d 3 r( E +c B ), () P = ǫ d 3 re B, () where the integration is over all space. Consider an expansion of the electric field in terms of plane waves: E( r,t) = E (π) 3 ( k,)ˆǫ ( ] k)e i( k r ωt) +c.c., (3) where ( k,) is a complex amplitude and c.c. stands for complex conjugate of the preceding term. The polarization vector satisfies: (a) Show that P can be written as P = ǫ c Note that all time dependence has canceled out. Explain. ˆǫ ( k) = ˆǫ ( k). (4) (π) 3 ( k,) ˆk. (5) Consider the Coulomb gauge, where A = cf. eq. (6.) of Jackson]. In the absence of external sources (ρ = J = ), we also have Φ = cf. eq. (6.3) of Jackson]. Using eq. (6.9) of Jackson, the electric and magnetic fields are given by, If we write where ω = kc and A( r,t) = E = A t, B = A. (6) a( (π) k)e i( k r ωt) + a ( ] k)e i( k r ωt), 3 a( k) = a ( k)ˆǫ ( k), then the vector amplitudes in the plane wave expansion of the electric and magnetic fields, obtained from eq. (6), are given by: ( k) = ikc a( k), B ( k) = i k a( k) = c ˆk ( k), (7)

2 where ( k) = ( k,)ˆǫ ( k). That is, B( r,t) = c E (π) 3 ( k,)ˆk ˆǫ ( ] k)e i( k r ωt) +c.c.. (8) Inserting eqs. (3) and (8) into eq. () taking care to employ different dummy variables in the sums and integrals], and expanding out the resulting expression, P = ǫ { d 3 r E (π) 6 ( c k,) ( k, )ˆǫ ( k) ˆk ˆǫ ( k )]e i( k+ k ) r e i(ω+ω )t +E ( k,)e ( k, )ˆǫ ( k) ˆk ˆǫ ( k )]e i( k+ k ) r e i(ω+ω )t + ( k,)e ( k, )ˆǫ ( k) ˆk ˆǫ ( k )]e i( k k ) r e i(ω ω )t +E ( k,) ( k, )ˆǫ ( k) ˆk ˆǫ ( k )]e i( k k ) r e i(ω ω )t }, (9) where ω kc and ω k c. In our notation, k k and k k. We may now perform the integral over r, using d 3 re i( k± k ) r = δ 3 ( (π) k± k ), () 3 and then use the delta function to integration over k. Then eq. (9) reduces to P = ǫ { E c (π) 3 ( k,) ( k, )ˆǫ ( k) ˆk ˆǫ ( k)]e iωt where we have used eq. (4) to write: E ( k,)e ( k, )ˆǫ ( k) ˆk ˆǫ ( k)]e iωt + ( k,)e( k, )ˆǫ ( k) ˆk ˆǫ ( k)] } +E ( k,) ( k, )ˆǫ ( k) ˆk ˆǫ ( k)], () ˆǫ ( k )δ 3 ( k+ k ) = ˆǫ ( k)δ 3 ( k+ k ) = ˆǫ ( k)δ 3 ( k+ k ). () We can now use the vector identity, ˆǫ ( k) ˆk ˆǫ ( k)] = ˆkˆǫ ( k) ˆǫ ( k)] ˆǫ ( k)ˆk ˆǫ ( k)] = ˆkδ, (3) Recall that for any well-behaved function f( k, k ) we have f( k, k )δ 3 ( k± k ) = f( k,± k)δ 3 ( k± k ), due to the presence of the delta function. For example, ω δ 3 ( k± k ) = k cδ 3 ( k± k ) = kcδ 3 ( k ± k ) = ωδ 3 ( k ± k ), since ± k = k.

3 using the properties of the polarization vector, ˆk ˆǫ ( k) =, and ˆǫ ( k) ˆǫ ( k) = δ. (4) Using eq. (3) allows us to perform the sum over in eq. (), which yields P = ǫ } { E c (π) ˆk 3 ( k,) ( k,)e iωt E ( k,)e ( k,)e iωt + ( k,). (5) Noting that ω = kc and ˆk k/k, it follows that ˆk ( k,) ( k,)e iωt =, since the integrand is an odd function under k k. Hence, eq. (5) yields P = ǫ c (π) ˆk E 3 ( k,), (6) which confirms the result of eq. (5). Note that P given in eq. (6) is explicitly time-independent. This is simply an expression of the conservation of momentum, d P/dt =. This is a consequence of eq. (6.) of Jackson. Since ρ = J = for a free electromagnetic field, we have P mech =, in which case d P dt = P field dt = S daˆn T=, where T isthemaxwell stresstensor. Theunitvector ˆnistheoutwardnormaltothesurfaceS, where S is the surface of infinity. For any finite energy field configuration, the stress tensor vanishes at the surface of infinity and we recover d P/dt = as expected. (b) Obtain the corresponding expression for the total energy U. Employing the photon interpretation for each mode ( k, ) of the electromagnetic field, justify the statement that photons are massless. The total energy is given (in SI units) by U = ǫ d 3 r( E +c B ). (7) We first compute E d 3 r = (π) 6 d r{ 3 ( k,) ( k, )ˆǫ ( k) ˆǫ ( ] k )e i( k+ k ) r e i(ω+ω )t +c.c. + ( k,)e ( k, )ˆǫ ( k) ˆǫ ( k )e i( k k ) r e +c.c.] } i(ω ω )t, 3

4 where the computation is similar to that of part (a). Integrating over r and using eq. () as we did in part (a), it follows that E d 3 r = { } E (π) 3 ( k,) ( k, )ˆǫ ( k) ˆǫ ( k)e iωt + ( k,)e ( k, )ˆǫ ( k) ˆǫ ( k)+c.c.. Summing over using eq. (4), we obtain E d 3 r = E (π) 3 ( k,) ( ] k,)e iωt +c.c. + (π) 3 ( k,). (8) Next, we compute c B d 3 r. The only difference in the computation compared to the one above is that ˆǫ ( k) is replaced by ˆk ˆǫ ( k) and ˆǫ ( k ) is replaced by ˆk ˆǫ ( k ). Thus, instead of obtaining the factor ˆǫ ( k) ˆǫ ( k )δ 3 ( k + k ) after the integration over r, we now have cf. footnote ]: ˆk ˆǫ ( k)] ˆk ˆǫ ( k )]δ 3 ( k+ k ) = ˆk ˆk ]ˆǫ ( k) ˆǫ ( k )] ˆk ˆǫ ( k )]ˆk ˆǫ ( k)]δ 3 ( k+ k ) = { ˆǫ ( k) ˆǫ ( k)+ˆk ˆǫ ( k)]ˆk ˆǫ ( k)] } δ 3 ( k+ k ) = ˆǫ ( k) ˆǫ ( k)δ 3 ( k+ k ), afterusing eqs. ()and(4). Similarly, instead ofobtaining thefactor ˆǫ ( k) ˆǫ ( k )δ 3 ( k k ) after the integration over r, we now have: Hence, it follows that: c B d 3 r = ˆk ˆǫ ( k)] ˆk ˆǫ ( k )]δ 3 ( k k ) = ˆǫ ( k) ˆǫ ( k)δ 3 ( k k ). Adding eqs. (8) and (9) yields U = ǫ E (π) 3 ( k,) ( ] k,)e iωt +c.c. + (π) 3 ( k,). (9) (π) 3 ( k,). () Note that U given in eq. () is explicitly time-independent. This is simply an expression of the conservation of momentum, du/dt =. This is a consequence of eq. (6.) of Jackson. Since ρ = J = for a free electromagnetic field, we have P mech =, in which case du dt = U field dt = S daˆn S =, where S is the Poynting vector. For any finite energy field configuration, the Poynting vector vanishes at the surface of infinity and we recover du/dt = as expected. 4

5 Finally, consider a fixed wave number vector k, for which ( k,) ()δ 3 ( k k ). Then, eqs. (6) and () yield U = ǫ (), P = ˆk ǫ c () = ˆk c U. That is, U = Pc. Comparing this result to the relativistic relation between the energy and momentum of a particle, E = p c +m c 4, we conclude that photons are massless.. Jackson, problem 7.4] A plane-polarized electromagnetic wave of frequency ω in free space is incident normally on the flat surface of a nonpermeable medium of conductivity σ and dielectric constant ǫ. (a) Calculate the amplitude and phase of the reflected wave relative to the incident wave for arbitrary σ and ǫ We take the wave to be traveling in the z-direction. Note that a plane-polarized wave is a synonym for a linearly polarized wave. Without loss of generality, we shall assume that the wave is linearly polarized in the x-direction. That is, we take the incident wave to be E = ˆxe i(kz ωt), and the reflected wave to be E = ˆxe i(kz+ωt). For normal incidence with the reflected wave polarized parallel to the plane of incidence (with µ = µ = µ ), eq. (7.4) of Jackson yields (in MKS units), = n n ǫ /ǫ = n +n ǫ /ǫ +, () where the index of refraction is n i = ǫi µ i ǫi =, for i =,. ǫ µ ǫ Since the incident plane wave propagates in free space, we identify ǫ = ǫ and µ = µ. By assumption, the reflecting surface is nonpermeable (i.e., µ = µ ). We can take the conductivity σ of the medium in region by writing cf. eq. (7.57) of Jackson]: ǫ = ǫ+ iσ ω, Hence, eq. () now becomes, E = ǫ + iσ ǫ ǫ ω ǫ + iσ. () ǫ ǫ ω + 5

6 To simplify this expression, we shall write ǫ ǫ + iσ ǫ ω = reiθ, where θ π (since ǫ, ǫ, σ and ω are all non-negative quantities). Taking the real part of this equation, we see that ǫ = rcosθ. (3) ǫ Furthermore, ( ) ( ) ǫ σ r = + = ǫ + σ ǫ ǫ ω ǫ ǫ ω. (4) It then follows that and We can therefore write Noting that we obtain ) / cosθ = (+ σ. ǫ ω = r/ e iθ/ r / e iθ/ +, (5) = (r/ e iθ/ )(r / e iθ/ ) (r / e iθ/ +)(r / e iθ/ +) = r + r/ cos(θ/) r ++r / cos(θ/). cos (θ/) = (+cosθ) = = r+ r++ ( + ǫ ), ǫ r ( r + ǫ with r given by eq. (4). To determine the complex number /E, we write To determine α, we use eq. (5) to write E = ǫ ) ( r + ǫ ), (6) ǫ eiα. E = (r/ e iθ/ )(r / e iθ/ +) (r / e iθ/ +)(r / e iθ/ +) = r +ir/ sin(θ/) r ++r / cos(θ/). It then follows that ) tanα = Im(E /E (r ǫǫ ) Re(/E ) = r/ sin(θ/) = r r after using sin (θ/) = ( cosθ) and making use of eq. (3)., 6

7 By definition, the reflected wave is given by E = ˆxe i(kz+ωt) = ˆxe i(α+π) e i(kz+ωt), after writing e iπ =. Hence, the amplitude of the reflected wave is r + r++ ( r + ǫ ǫ ( r+ ǫ ǫ ) ) and the relative phase of the incident and reflected wave at the interface (z = ) is /, ( ) (r φ rel α+π = π +tan ǫ/ǫ ), (7) r where r = ǫ + σ ǫ ǫ ω. (8) (b) Discuss the limiting cases of a very poor and a very good conductor, and show that for a good conductor the reflection coefficient (ratio of reflected to incident intensity) is approximately R ω δ, (9) c where δ is the skin depth. First, we discuss the poor conducting limit, σ ǫω. It is convenient to introduce the dimensionless ratio, σ ǫω. Then, eq. (8) yields r ǫ ( + ǫ, Using eq. (6), we identify the reflection coefficient as ( R ǫ ǫ (+ ǫ )+ ǫ + ) 4 ( = ( ǫ/ǫ ) + ] ǫ/ǫ ǫ/ǫ ǫ ǫ (+ ǫ )++ ǫ + ) ( ǫ/ǫ +) + ]. ǫ/ǫ 4 ǫ/ǫ + We are assuming that ǫ > ǫ in which case r > and the complex number E / lies in the first quadrant of the complex plane where both its real and imaginary parts are positive. In this case < α < π. 7

8 Expanding out the denominator, we obtain after some algebra, ( ) R ] ǫ/ǫ ǫ = 3ǫ/ǫ ǫ/ǫ + ǫ (ǫ/ǫ ) +O(4 ). Similarly, eq. (7) yields φ rel π + ǫ/ǫ ǫ/ǫ +O( ), under the assumption that (ǫ/ǫ ). 3 Next, we consider the limit of a good conduction, corresponding to σ ωǫ. In this limit, r σ ǫ ω. Hence, eq. (6) yields R Introducing the skin depth, we see that (using µ = µ ) that R r r r+ r /r + /r r ǫ ω σ. δ ωσµ, (3) ωδ ǫ µ ωδ c after making use of the relation ǫ µ = /c. Employing the same approximation (r ), it follows that φ rel π +tan r π + r π + ω c δ., (3) 3. Jackson, problem 7.6] A plane wave of frequency ω is incident normally from vacuum on a semi-infinite slab of material with a complex index of refraction n(ω) n (ω) = ǫ(ω)/ǫ ]. (a) Show that the ratio of the reflected power to the incident power is R = n(ω) +n(ω), (3) while the ratio of power transmitted into the medium to the incident power is T = 4Ren(ω) +n(ω). (33) 3 By assumption of the problem, ǫ ǫ. So we see how small must be in order for the first order approximation of the arctangent to be valid. 8

9 The derivation of the Fresnel equations holds for complex index of refraction, so we conclude that for a wave incident normally to the plane (with µ = µ ), eq. (7.4) of Jackson gives, ( ) n(ω) = ±, n(ω)+ where the plus (minus) sign refers to the incident wave polarized parallel (perpendicular) to the plane of incidence. As in the previous problem, we shall assume that the incident wave is traveling in the z-direction and is linearly polarized in the x-direction. Then, the incident and the reflected waves are given by, E = ˆx e i(kz ωt), E = ˆxE e i(kz+ωt). The corresponding magnetic fields are obtained using B = ǫ µ ˆk E, B = ǫ µ ˆk E, where ˆk = ẑ and ˆk = ẑ for normal incidence. Hence, B = ŷ c ei(kz ωt), B = ŷ E c e i(kz+ωt), after using ǫ µ = /c. The (complex) Poynting vector points in the z-direction, S z = Re( E H ) z = ǫ, µ after using B = µ H. Similarly, S z = ǫ. µ Hence, it follows that R = S z S z = which is equivalent to eq. (3). Next, we consider the transmitted waves, = n(ω) n(ω)+, E = ˆxE ei(k z ωt), (34) B = ŷ ǫµ E e i(k z ωt), (35) where k = kn(ω) = (ω/c)n(ω). 9

10 For normal incidence (with µ = µ ), eq. (7.4) of Jackson gives, E = n(ω)+. The (complex) Poynting vector at the interface (i.e., at z = ) points in the z-direction, S z = Re( E H ) z = E Re ǫ µ = E Ren(ω), (36) after making use of, Thus, it follows that T = S z S z n(ω) = = ǫµ ǫ =. ǫ µ ǫ in agreement with eq. (33). Note that as expected, Ren(ω) = 4Ren(ω) n(ω)+, (37) T +R = n(ω) +4Ren(ω) n(ω)+ =. (b) Evaluate Re iω E D B H ) ] as a function of (x,y,z). Show that this rate of change of energy per unit volume accounts for the relative transmitted power T. Using D = ǫ E, it then follows (for µ = µ ) that, E D B H = ǫ E E µ B B = ǫ E e i(k k )z ǫǫ E e i(k k )z, where the factors of µ have cancelled. We next observe that ǫǫ = ǫ = ǫ n(ω) 4 and where k = ω/c. Hence, Noting that k k = k n(ω) n (ω) ] = ikimn(ω), (38) E D B H = ( n (ω)] n(ω) ) ǫ E e kzimn(ω). (39) n (ω)] n(ω) = Ren iimn] (Ren) (Imn) = i(ren)(imn) (Imn), it follows that Re iω( E D B H ) ] = ωǫ (Ren)(Imn)e kzimn(ω). (4)

11 The complex Poynting theorem for harmonic fields states that for J = cf. eq. (6.34) of Jackson], iω ( E D B H ) d 3 x = S ˆnda, V S where S = ( E H ) is the complex Poynting vector and ˆn points out of the volume V. If we decompose d 3 x = dadz, where da is the infinitesimal area element transverse to the z-direction, and identify ˆn = ẑ, then we can write, Re iω( E D B H ) ] dz = ReS z. (4) We interpret Re iω( E D B H ) ] as the rate of change of energy per unit volume. Using eq. (4), Re iω( E D B H ) ] dz = ωǫ E (Ren)(Imn) e kzimn(ω) dz = ǫ c E Ren, after using k = ω/c. Physically, this quantity is the transmitted energy flux (i.e. energy per unit time per unit area). The transmission coefficient T is simply the ratio of the transmitted energy flux to the incident energy flux. In part (a), we made use of eq. (7.3) of Jackson, which states that the incident energy flux (which Jackson calls the energy flow) is given by, S ˆn = ǫ µ. Hence we can identify T = ǫ c E Ren ǫ /µ. Finally, using ǫ µ = /c, we end up with in agreement with eq. (37). T = Ren(ω), (c) For a conductor with n = +i(σ/ωǫ ), where σ is real, write out the results of parts (a) and (b) in the limit of ǫ ω σ. Express your answer in terms of δ as much as possible. Calculate Re( J E) and compare with the results of part (b). Do both enter the complex form of Poynting s theorem? Using n (ω) = + iσ ωǫ, (4)

12 under the assumption that ǫ ω σ, we insert eq. (4) into eq. () and obtain, R = + iσ = ωǫ + iσ. + ωǫ This result precisely matches the result obtained in problem (a) after setting ǫ = ǫ in eq. (). Thu, we can make immediate use of eq. (3) to obtain. R ωδ c where the skin depth δ is defined in eq. (3). Moreover, since T +R =, it follows that T ω c δ. In class, we noted that in the limit of σ ǫ ω, we have k = n(ω) ω c +i δ It then follows that Ren(ω) = Imn(ω) c ωδ. (43) Inserting this result into eq. (39) then yields, Re iω( E D B H ] = ǫ c e z/δ, ωδ after using k = ω/c. The expression above is further simplified after using eq. (3) and c = /(ǫ µ ), Re iω( E D B H ) ] = σ E e z/δ. (44) Next, we use J = σ E and compute, Re J E = σre E E = σ E Ree i(k k )z = σ E e z/δ (45) after using eqs. (34) and (38) and expressing the result in terms of the skin depth cf. eq. (3)]. The complex Poynting theorem see eq. (6.34) of Jackson] states that., J E + S + iω( E D B H ) =. (46) We therefore compute cf. eq. (36)], Re S = Re S z z = ǫ E Ren(ω) µ z e z/δ. (47)

13 In light of eq. (43), Ren(ω) z e z/δ = c ωδ = µ cσ, after using eq. (3). Inserting this result back into eq. (47) and putting ǫ µ = /c, we end up with Re S = σ E e zδ. (48) Adding up eqs. (45) and (48) yields Re J E + S ] =, which cannot be consistent with the complex Poynting theorem given in eq. (46) in light of eq. (44). The resolution of this paradox is as follows. Returning to the complex Poynting theorem given by eq. (46), there are two equivalent methods for treating the conducting medium.. D = ǫ E and J = σ E or (. D = ǫ + iσ ) E and J =. ω In method (), the conduction current and charges are designated as free, whereas in method () they are designated as bound (and therefore incorporated into D). The paradox encountered above arises because we used both D = (ǫ + iσ/ω) E and J = σ E simultaneously (which double counts the effect of the conduction current in the energy budget). In method (), the Re J E in eq. (46) is absent. Using eqs. (44) and (48), which we rewrite below, we see that Re iω( E D B H ) ] = σ E e z/δ, Re S = σ E e zδ, (49) Re S + iω( E D B H ) ] =, (5) which is consistent with the complex Poynting theorem cf. eq. (46)] with J =. Note that in method (), we can write iω E D = iωǫ E E +σ E E = iωǫ E E + J E, after identifying J = σ E in the last term above. Thus, we can rewrite eq. (5) as Re J E + S + iω( E (ǫ E ) B H )] =. This result is consistent with the complex Poynting theorem applied to method (), where we identify D = ǫ E. We conclude that in both methods () and () for treating a conducting medium, the complex Poynting theorem is indeed satisfied. 3

14 4. Jackson, problem 7.7] The angular momentum of a distribution of electromagnetic fields in vacuum (in SI units) is given by L = d 3 r r ( E B), (5) µ c where the integration is over all space. (a) For fields produced a finite time in the past (and so localized to a finite region of space) show that, provided the magnetic field is eliminated in favor of the vector potential A, the angular momentum can be written in the form L = µ c d 3 r E A+ ] 3 E l ( r )A l. (5) The first term above is sometimes identified with the spin of thephoton and the second with the orbital angular momentum because of the presence of the angular momentum operator L op = i( r ). l= The magnetic field can be written in terms of the vector potential, B = A. Hence, we need to evaluate r E ( A)]. Using the Einstein summation convention, where there is an implicit summation over a pair of identical indices, we can write 4 ( a b) i = ǫ ijk a j b k, where the indices take on the values i,j,k =,,3 and there is an implicit sum over j and k. The Levi-Civita tensor is defined as +, if (i,j,k) is an even permutation of (,,3), ǫ ijk =, if (i,j,k) is an odd permutation of (,,3),, otherwise. Thus, it follows that { r E ( A)] } i = ǫ ijkx j E ( A)] k = ǫ ijk x j ǫ klm E l ( A) m = ǫ ijk x j ǫ klm E l ǫ mpq p A q, where r (x,x,x 3 ) and p / x p. We now employ the following ǫ-identity, ǫ klm ǫ mpq = δ kp δ lq δ kq δ lp. Hence, it follows that { r E ( A)] } i = ǫ ijkx j E l (δ kp δ lq δ kq δ lp )δ lp = ǫ ijk x j E l k A l ǫ ijk x j E l l A k. (53) 4 In this calculation, all vectors of Euclidean three-vectors. Consequently, we do not distinguish between upper and lower indices. All indices will be written as subscripts in what follows. 4

15 We recognize ǫ ijk x j E l k A l = E l ( r ) i A l which corresponds to the second term in eq. (5). To obtain the first term in eq. (5) will require an integration by parts. That is, we first write: ǫ ijk x j E l l A k = ǫ ijk l (x j E l A k ) A k l (x j E k )], which is an identity that follows from the rule for differentiating products. Next, we note that ǫ ijk A k l (x j E l ) = ǫ ijk A k x j ( l E l )+E l ( l x j )] = ǫ ijk A k E l δ lj = ǫ ijk A k E j = ( E A) i, where we used δ l x j x j / x l = δ lj and l E l = E = (in vacuum). Thus, eq. (53) yields the vector identity, { r E ( A)] } i = E l( r ) i A l +( E A) i ǫ ijk l (x j E l A k ), (54) where there is an implicit sum over the repeated index l. When we integrate over all of space, we can use the divergence theorem given in the inside cover of Jackson s textbook]: d 3 rǫ ijk l (x j E l A k ) = daǫ ijk n l x j E l A k = daˆn E( r A) i =, (55) V S where n l is theoutward normal at the surface S ands is the surface ofinfinity. Since the fields are assumed to be localized to a finite region of space, the integral above vanishes. Hence, inserting the results of eqs. (54) and (55) into eq. (5) after putting B = A] immediately yields ] 3 d 3 r r ( E B) = d 3 r E A+ E l ( r )A l. Therefore, eq. (5) is proven. REMARK: The identification of L spin = µ c S l= d 3 r E A, (56) as the spin angular momentum is problematical, as eq. (56) is not invariant under gauge transformations. In fact, a gauge-invariant expression for the spin angular momentum can be constructed that reduces to eq. (56) in the radiation (Coulomb) gauge. 5 (b) Consider an expansion of the vector potential in the radiation (Coulomb) gauge in terms of plane waves, A( r,t) = ˆǫ (π) 3 ( k)a ( ] k)e i( k r iωt) +c.c.. (57) 5 See e.g., Iwo Bialynicki-Birula and Zofia Bialynicki-Birula, Journal of Optics 3, 644 () and references therein. 5

16 The vectors ˆǫ ( k) are conveniently chosen as the positive and negative helicity polarization vectors 6 ˆǫ ± = (ˆǫ ±iˆǫ ), (58) where ˆǫ and ˆǫ are the real orthogonal vectors in the plane whose positive normal is in the direction of k. Show that the time average of the first (spin) term of L can be written as L spin = µ c (π) 3 k a + ( k) a ( ] k). Can the term spin angular momentum be justified from this expression? Calculate the energy of the field in terms of the plane wave expansion of A and compare. In the Coulomb gauge, the electric field is (in SI units): E( r,t) = A t = i (π) ω ˆǫ 3 ( k)a ( ] k)e i( k r iωt) c.c., (59) where ω = ck and k k. Note that due to the overall factor of i, we must subtract the complex conjugate inside the square brackets in order to ensure that E( r,t) is a real field. Inserting eqs. (57) and (59) into eq. (56) and expanding out the integrand, we obtain: L spin = µ c i (π) 6 { ω d 3 r ˆǫ ( k) ˆǫ ( k )]a ( k)a ( k )e i( k+ k ) r e i(ω+ω )t +ˆǫ ( k) ˆǫ ( k )]a ( k)a ( k )e i( k k ) r e i(ω ω )t ˆǫ ( k) ˆǫ ( k )]a ( k)a ( k )e i( k k ) r e i(ω ω )t ˆǫ ( k) ˆǫ ( k )]a ( k)a ( k )e i( k+ k ) r e i(ω+ω )t where ω = kc and ω = k c. We may now perform the integral over r, using eq. (), and then use the delta function to integrate over k. The end result is L spin = i µ c ω (π) 3 { ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k) ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k) +ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k)e iωt ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k)e iωt }. 6 Jackson omits the overall factor of in the definition of ˆǫ ±. I prefer to maintain this phase convention, but you are free to choose any convention that suits you. }, (6) 6

17 However, the last two terms above vanish when integrated over k, since the corresponding integrands are odd functions of k. For example, under k k, ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k)e iωt ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k)e iωt, = ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k)e iωt, = ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k)e iωt, where we interchanged and in the penultimate step (which is justified since these are dummy labels that are being summed over), and used the antisymmetry of the cross product in the final step. Note that ω = k c does not change sign when k k. Hence, eq. (6) simplifies to L spin = i µ c ω (π) 3 { } ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k) ˆǫ ( k) ˆǫ ( k)]a ( k)a ( k). (6) Using the definition of the polarization vectors given in eq. (58), it is straightforward to verify that 7 ˆǫ ( k) ˆǫ ( k) = iˆkδ, for, = ±. (6) This result allows us to sum over in eq. (6). Both terms in eq. (6) contribute equally and the end result is: L spin = µ c (π) 3 k { a + ( k) a ( k) }, (63) after using ω = kc and k = kˆk. Note that L spin is time-independent and thus conserved. This is a stronger condition than the conservation of angular momentum, which only requires that the sum L = L orbital + L spin is conserved. Eq. (63) implies that the spin angular momentum of the electromagnetic field is separately a constant of the motion. 8 If we interpret each mode ( k,) as a photon, then the two possible photon spin states (in a spherical basis) correspond to positive and negative helicity, i.e. states of definite spin angular momentum in which L spin points in a direction parallel or antiparallel to the direction of propagation ˆk, respectively. 7 To prove eq. (6), use the fact that ˆǫ ˆǫ = ˆǫ ˆǫ = ˆk and ˆǫ ˆǫ = ˆǫ ˆǫ =. 8 Indeed, Jackson only asks that we show that the time-average of L spin is given by eq. (63). In such a calculation, the last two terms in eq. (6) are immediately set to zero when taking the time-average since the time-averaged values e ±iωt = T T e ±iωt dt =, when ω, where T = π/ω is the time for one oscillation cycle. The case of ω = corresponds to k =, in which case the last two terms in eq. (6), when summed over and, are each manifestly equal to zero, since eq. (58) implies that ˆǫ ( k) ˆǫ ( k) = for = ± (and the cross-terms vanish). However, our result above is more general since no time-averaging is required to obtain eq. (63). 7

18 It is instructive to consider the energy of the electromagnetic fields, which was obtained in problem. In particular, eq. () yields U = ǫ (π) 3 ω a ( k), (64) where we have used eq. (7) to write ( k,) = iωa ( k). Consider a fixed mode of positive helicity ( k, = +). Then, a ( k) = a + ( k )δ( k k )δ,+, in which case eq. (64) yields and U = ǫ ω (π) 3 a ( k ), L spin = µ c (π) 3 k a ( k ) = ǫ ω (π) ˆk 3 a ( k ), after using ǫ µ = /c and k = (ω /c)ˆk. That is, L spin = U ω ˆk, for = +. (65) For a fixed mode of negative helicity ( k, = ), we again obtain eq. (65) with =. For a single photon of frequency ω, quantum mechanics states that U = ω, and eq. (65) yields L spin = ± ˆk, corresponding to a spin-one particle of helicity ±, with its spin parallel or antiparallel to the direction of propagation ˆk. 5. (a) Assume that the vector potential in the Lorentz gauge is given by: A( x, t) = A (x,y)(ˆx±iŷ)e i(kz ωt), (66) where A (x,y) is a very slowly varying function of position. Slowly varying means that the second spatial derivatives of A (x,y) can be neglected; however, one must not neglect first derivatives of A (x,y). Derive the approximate forms for the electric and magnetic fields given in Jackson, problem 7.8, E(x,y,z,t) (x,y)(ˆx±iŷ)+ i ( ] E k x ±i )ê 3 e ikz iωt, (67) y B(x,y,z,t) i µǫ E(x,y,z,t), (68) where ˆx, ŷ and ẑ are unit vectors in the x, y and z directions, respectively. The Lorenz gauge condition (in MKS units) is A+ c Φ t =. 8

19 Using eq. (66), Integrating, we get ( ) A A = x ±i A e ikz iωt = Φ y c t =. Φ( x, t) = ic ω ( ) A x ±i A e ikz iωt, (69) y where we have dropped the integration constant without loss of generality, since the scalar potential is only defined up to an additive constant. The electric and magnetic fields are given by E = Φ A t, B = A. Plugging in eqs. (66) and (69), Φ kc ω ( ) A x ±i A ẑe ikz iωt, y where we have dropped second spacial derivatives of A and A t = iωa (ˆx±iŷ)e ikz iωt, If we define (x,y) iωa (x,y) and make use of ω = ck, we end up with, E(x,y,z,t) (x,y)(ˆx±iŷ)+ i ( ) ] E k x ±i ẑ e ikz iωt. y Next, we evaluate ˆx ŷ ẑ B(x,y,z,t) = A = det x y z A e ikz iωt ±ia e ikz iωt ( ) ] A = i (ˆx±iŷ)ikA (x,y) x ±i A ẑ e ikz iωt, y = i ǫ µ E(x,y,z,t), (7) after using (x,y) iωa (x,y) = icka (x,y) and ǫ µ = /c. Finally if the electromagnetic waves are propagating in a simple nonconducting medium where D = ǫ E and B = µ H, the eq. (7) is modified by simply replacing ǫ and µ with ǫ and µ, respectively. 9

20 (b) Jackson, problem 7.9] For the circularly polarized wave given by eqs. (67) and (68), with (x,y) a real function of x and y, calculate the time-averaged component of the angular momentum parallel to the direction of propagation. Show that the ratio of this component of angular momentum to the energy of the wave in vacuum is, L 3 U = ±ω. Interpret this result in terms of quanta of radiation (photons). Show that for a cylindrically symmetric, finite plane wave, the transverse components of angular momentum vanish. The angular momentum density of the electromagnetic field is given by cf. problem 6. on p. 88 of Jackson]. L = x g = ǫµ x ( E H). Using the vector identity, x ( E H) = E( x H) H( x E), The z component of the angular momentum density (denoted below by L 3 ) is given by L 3 = ǫµ x(e z H x E x H z )+y(e z H y E y H z ) ]. (7) The results quoted above assume that E and H are real fields. Thus, taking the real part of the fields given in eqs. (67) and (68), we have E x = cos(kz ωt), E y = sin(kz ωt), (7) E z = i E k x sin(kz ωt)± E ] cos(kz ωt) (73) y ǫ H x = ± µ sin(kz ωt), H y = cos(kz ωt), (74) H z = i ǫ ± k µ x cos(kz ωt) E ] sin(kz ωt), (75) y after using B = µ H. Inserting the above results into eq. (7) yields. L 3 = ǫ ] ǫµ x k x +ye. (76) y In the medium, we have k = ǫµω. Thus, we can rewrite eq. (76) as L 3 = ǫ x ω x +y ] E. (77) y Next, we compute the energy density cf. eq. (6.6) of Jackson], u = (ǫ E +µ H ).

21 As in part (a), we shall assume that (x,y) is slowly varying so that we can neglect the second spatial derivatives of. That is we may discard terms proportional to ( / x), ( / y) and ( / x)( / y) as compared to terms proportional to E. In particular, in evaluating E and H, we can drop the contributions from E z and H z. Hence, u ǫe +µ ( ǫ µ E )] = ǫe. Finally, we compute the total energy and the z-component of the total angular momentum, U = d 3 xu = ǫ d 3 xe (x,y), L 3 = d 3 L 3 = ǫ ω d 3 x x x E +y ] y E = ǫ ω d 3 xe (x.y), after integrating by parts and using the fact that (x,y) vanishes when x, y. We conclude that L 3 U = ± ω. The interpretation in terms of the photon is clear. Since a photon has an energy U = ω, it follows that L 3 = ± for the photon. The two possible signs correspond to positive and negative helicity. To complete the problem, we compute the x and y components of the angular momentum density (denoted below by L and L ). L = ǫµ y(e x H y E y H x )+z(e x H z E z H x ) ], L = ǫµ x(e y H x E x H y )+z(e y H z E z H y ) ]. Inserting the fields given in eqs. (7) (75), we end up with ǫ L = ǫµ µ E y ± ǫ k µ z E ], x ǫ L = ǫµ µ E x± ǫ k µ z E ]. y By assumption, the plane wave is cylindrically symmetric, which implies that Thus, (x,y) = ( x,y), ǫ L = ǫµ µ (x,y) = (x, y). d 3 x ye(x,y)± z ] k x E (x,y) =, since the integrand is an odd function under x x, y y. Likewise, ǫ L = ǫµ d 3 x xe µ (x,y)± z ] k y E (x,y) =. Hence, we conclude that for a cylindrically symmetric finite circular polarized electromagnetic wave, L = L = and L 3 = ±U/ω.